A hemispherical bowl of radius 10 cm is placed in a uniform electric field of \(1000\ \mathrm{N\,C^{-1}}\). The field is parallel to the axis of the bowl, and there is no charge in the region. The magnitude of the electric flux through the curved surface of the bowl is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(31.4\ \mathrm{N\,m^{2}\,C^{-1}}\)
Close the bowl with a flat circular lid across its rim. The lid and the curved surface together form a closed surface with no charge inside. By Gauss’s law the total flux through it is zero. So the flux through the curved surface is equal in size to the flux through the lid.
The lid is normal to the field, so the flux through it is \( E\times\pi R^{2} \).
\( \begin{aligned} \phi &= E\,\pi R^{2} \\ &= 1000\times3.14\times(0.1)^{2} \\ &= 31.4\ \mathrm{N\,m^{2}\,C^{-1}} \end{aligned} \)
NCERT (2026-27):
Gauss’s Law – “The law implies that the total electric flux through a closed surface is zero if no charge is enclosed by the surface.”